Question

What is the pH of a solution prepared by mixing 60.0 mL of 0.20M potassium carbonate (K2CO3) with 140.0 mL of 0.1M potassium bicarbonate (KHCO3)? a. 6.43 b. 6.30 c. 6.67 d. 10.32 e. 10.18

Answer 1

Answer – Given, [K₂CO₃] = [CO₃^2-] = 0.20 M

[KHCO₃] = [HCO₃^-] = 0.10 M

Moles of CO₃^2- = 0.20 M * 0.060 L

                         = 0.012 moles

Moles of HCO₃^- = 0.10 M * 0.140 L

                            = 0.014 moles

New concentration

Total volume = 140+60 = 200 mL

[CO₃^2-] = 0.012 moles / 0.200 L = 0.06 M

[HCO₃^-] = 0.0140 / 0.200 L = 0.07 M

We know the pKa2 for H₂CO₃ = 10.25

We know the Henderson Hasselbalch equation –

pH = pKa2 + log [CO₃^2-] / [HCO₃^-]

     = 10.25 + log 0.06 / 0.07

     = 10.18

So the pH of a solution prepared by mixing 60.0 mL of 0.20M potassium carbonate (K2CO3) with 140.0 mL of 0.1M potassium bicarbonate (KHCO3) is e ) 10.18